[C++]LeetCode: 50 Majority Element

Given an array of size n, find the majority element. The majority element is the element that appears more than ? n/2 ? times.

You may assume that the array is non-empty and the majority element always exist in the array.

找出数组中出现次数超过数组长度一半的值。

Solution:

1. Runtime: O(n2)— Brute force solution: Check each element if it is the majority element.
2. Runtime: O(n), Space: O(n)— Hash table: Maintain a hash table of the counts of each element, then find the most common one.
   1. Runtime: O(n log n)— Divide and conquer: Divide the array into two halves, then find the majority element A in the first half and the majority element B in the second half. The global majority element must either be A or B. If A == B, then it automatically becomes the global majority element. If not, then both A and B are the candidates for the majority element, and it is suffice to check the count of occurrences for at most two candidates. The runtime complexity, T(n) = T(n/2) + 2n = O(n logn).
   2. Runtime: O(n) —Moore voting algorithm: We maintain a current candidate and a counter initialized to 0. As we iterate the array, we look at the current element x:
      
      1. If the counter is 0, we set the current candidate to x and the counter to 1.
      2. If the counter is not 0, we increment or decrement the counter based on whether x is the current candidate. After one pass, the current candidate is the majority element. Runtime complexity = O(n)
         Moore Voting Algorithm: 该算法要求目标数组存在majority元素（大于n/2），否则需要检验。 算法演示 here. 思路解析： 1. 初始化majorityIndex，并且维护其对应count； 2. 遍历数组，如果下一个元素和当前候选元素相同，count加1，否则count减1； 3. 如果count为0时，则更改候选元素，并且重置count为1； 4. 返回A[majorityIndex] 原理：如果majority元素存在（majority元素个数大于n/2,个数超过数组长度一半），那么无论它的各个元素位置是如何分布的，其count经过抵消和增加后，最后一定是大于等于1的。 如果不能保证majority存在，需要检验。 复杂度：O(N) Attention: 循环时从i = 1开始，从下一个元素开始，因为count已经置1. AC Code:

         class Solution {
         public:
             int majorityElement(vector &num) {
                 //the majority element 存在并且唯一
                 
                 int majorityIndex = 0;
                 for(int cnt = 1, i = 1; i < num.size(); i++)
                 {
                     num[majorityIndex] == num[i] ? cnt++ : cnt--;
                     if(cnt == 0)
                     {
                         cnt = 1;
                         majorityIndex = i;
                     }
                 }
                 
                 return num[majorityIndex];
             }
         };

         
         检验： /* Function to check if the candidate occurs more than n/2 times */ boolisMajority(inta[], intsize, intcand) { inti, count = 0; for(i = 0; i < size; i++) if(a[i] == cand) count++; if(count > size/2) return1; else return0; }